The same logic from this solution to a special case applies here, except that you disregard the comments about $F_2$ and $F_3$ and just settle for all the quotients by prime ideals being fields.
The battle plan is, briefly:
- The intersection of all prime ideals is the zero ideal
- The quotient by any prime ideal is in fact a field.
- The ring embeds into a product of quotient rings of the form $R/P$ where $P$ is a prime ideal. (And of course, that is a product of fields.)
So such a ring is a subring of a product of fields, and is not necessarily the whole product, or a finite product.
If the ring is finite, or better yet, only has finitely many maximal ideals, then the injection above is into a product of only finitely many fields. A collection of maximal ideals is always comaximal, so the Chinese Remainder Theorem says the map is surjective, and so it is actually an isomorphism.